You already wrote down everything you need.
First you wrote:
$$x H_n = \frac{1}{2}H_{n+1} + n H_{n-1}$$
Then you wrote:
$$I = \frac{1}{\alpha^2}\int_{-\infty}^\infty H_n(\mu) H_m(\mu) e^{-\mu^2} \cdot \mu \cdot d\mu$$
So.. now you can substitute $\mu H_n$:
$$I = \frac{1}{\alpha^2}\int_{-\infty}^\infty e^{-\mu^2} H_m(\mu) \left[\frac{1}{2}H_{n+1}(\mu) + n H_{n-1}(\mu)\right] d\mu$$
Where you can use now something you wrote already:
$$\int_{-\infty}^\infty e^{-x^2}H_n(x) H_m(x) dx = \sqrt{\pi} 2^n n! \delta_{nm}$$
Does this make it easier?
If you wish to use Dirac notation then you should not put operators inside either bra or ket symbols. You can write
$$
\hat{Q} | \phi \rangle
$$
and
$$
\langle \psi | \hat{Q} | \phi \rangle
$$
and
$$
( \hat{Q} | \phi \rangle )^\dagger = \langle \phi | \hat{Q}^\dagger
$$
but the notation
$$
\langle \hat{Q}^\dagger \phi | \;\;\;\;\;\;\;\;\;????
$$
is an abuse of Dirac notation. Of course you can write it if you want to adopt some other notation: that is up to you. You could, for example, announce that this quantity shall be defined, in your own work, to be identical to
$$
\langle \phi | \hat{Q}
$$
which appears to be what you are doing in your opening formula. But in Dirac notation you should not confuse matters by putting operators inside bra symbols or ket symbols. Inside the bra or ket should be simply a label: a symbol to say which bra or ket it is. The operator then acts on the whole thing. That is, $\hat{Q}$ acts on the ket $| \psi \rangle$ not on the label $\psi$.
You may be unfamiliar with writing an operator on the right of a bra, or as the last part of an expression. But it is in fact entirely logical and maps correctly to the mathematical behaviour of these mathematical objects. If in doubt, consider the case where an operator can be represented by a matrix and a ket can be represented by a vector.
[Final remark to pre-empt comments disagreeing with the above.
Mathematical notation does not have to be Dirac notation, so if you have seen something like $| \hat{Q} \psi \rangle$ or $\langle \hat{Q} \psi|$ in a reputable text, it does not mean I am saying that text is wrong, I am just saying that at this point it departs from Dirac notation, and I would not recommend someone learning the subject to depart from well-formed Dirac notation.]
Further note
A much-used fact about kets and their decomposition is
$$
| \psi \rangle = \sum_n | n \rangle \langle n | \psi \rangle
$$
where $\{ |n \rangle\}$ is a complete orthonormal set of kets (also called state-vectors). By repeatedly using this one finds things like
$$
\langle \phi | \hat{Q} | \psi \rangle
= \sum_n \sum_m
\langle \phi | n \rangle \langle n | \hat{Q}
| m \rangle \langle m | \psi \rangle .
$$
It follows that if $v_\phi$ is the vector whose components are $\langle n | \phi \rangle$ and $v_\psi$ is the vector whose components are $\langle n | \psi \rangle$ and
$Q$ is the matrix whose components are $\langle n | \hat{Q} | m \rangle$ then
$$
\langle \phi | \hat{Q} | \psi \rangle
= v_\phi^\dagger Q v_\psi
$$
where the expression on the right is written in the standard way for products of vectors and matrices.
For any such vectors and matrix it is the case that
$$
v_\phi^\dagger Q v_\psi
= (Q^\dagger v_\phi)^\dagger v_\psi
$$
and I expect this explains why someone was proposing to move the $Q$ operator in front of the $\phi$ and put a dagger on it in the incorrectly-formed expression which the question asked about.
A similar fact arises for continuous variables.
In the case of a continuous set of states the sums become integrals, and if $| x \rangle$ is a position eigenstate then
the quantity $\langle x | \psi \rangle$ is commonly written $\psi(x)$ and called a wavefunction. Here the letter $\psi$ is serving first as a label on a ket, and then as the name of a function. In this case the above expression becomes, for example,
$$
\langle \phi | \hat{Q} | \psi \rangle
= \int_{-\infty}^\infty \phi^*(x) Q \psi(x) dx
$$
where $Q$ is now an operator acting on functions of $x$. And we can, if we wish, now use
$$
\langle \phi | \hat{Q} | \psi \rangle
= \int_{-\infty}^\infty (Q^\dagger \phi(x))^\dagger \psi(x) dx.
$$
One can thus regard an operator either as acting to the right or as its adjoint acting to the left.
Best Answer
$\renewcommand\mean[1]{\langle #1\rangle}$ $\renewcommand\norm[1]{||#1||}$ $\renewcommand\h{\hbar}$ $\renewcommand\ket[1]{|#1\rangle}$ $\renewcommand\expval[2]{\langle #1|#2|#1\rangle}$ $\renewcommand\braket[2]{\langle #1|#2\rangle}$ $\renewcommand\Braket[3]{\langle #1|#2|#3\rangle}$ OP's original question
has already been answered in the replies of @Cosmas and @Zack. I won't go over these again. Instead, I propose here an alternative method of finding the ground state of the Harmonic Oscillator, using this time the concept of coherent states. I am going to assume the reader knows nothing about them, so I will build up everything here to avoid losing anyone.
Introduction to coherent states
There are three different - equivalent - ways to define a coherent state. For the sake of answering your question, I will only need one of them (sadly the less intuitive of them all) so I won't go all the way to introduce them all. The plan is thus the following: defining the so-called Heisenberg Uncertainty Principle and using it to define a coherent state. From there, I will be able to build a mathematical structure that will allow us to find the ground state of the Quantum Harmonic Oscillator.
Theorem 1 (Heisenberg Uncertainty Principle) For any operators $\hat{A}$ and $\hat{B}$, we have that $\Delta\hat{A}\hat{B}\geq\frac{1}{2}\norm{\mean{[\hat{A},\hat{B}]}}$.
In the case of the position and momentum operators, one has that $[\hat{X},\hat{P}] = i\h$. Applying this to the uncertainty principle, we find the famous relation $\Delta\hat{X}\Delta\hat{P}\geq \frac{\h}{2}$.
Definition 2 (Coherent states) A coherent state is defined as a state for which the Heisenberg Uncertainty Principle (applied to the operators $\hat{X}$ and $\hat{P}$ in the context of your question) is saturated, such that $\Delta\hat{X}\Delta\hat{P}=\frac{\h}{2}$.
Wave function of a coherent state
My goal here is to construct a wave function that depicts how a coherent state behaves. Let us introduce $\mathcal{O}(\lambda)=f^\dagger(\lambda)f(\lambda)$, where we define $f(\lambda) = \hat{A'}+i\lambda\hat{B'}, \hat{A'} = \hat{A}-\mean{\hat{A}}$ and $\hat{B'} = \hat{B}-\mean{\hat{B}}$.
Let us observe that the mean of $\mathcal{O}$ is positive. Indeed, for any $\ket{\psi}\in\mathcal{H}$, one has that \begin{equation} \expval{\psi}{f^\dagger(\lambda)f(\lambda)} = \norm{f(\lambda)\ket{\psi}}^2 \geq 0 \end{equation}
However, one also has that \begin{equation}\tag{1}\label{mean value O} \mean{\mathcal{O}(\lambda)} = \lambda^2\mean{\hat{B'}}+\lambda\mean{i[\hat{A'},\hat{B'}]}+\mean{\hat{A'}^2}. \end{equation} Notice that this is a quadratic equation in $\lambda$. We can compute its minimum through derivation. One finds that $\lambda_{min} = -\frac{\mean{i[\hat{A'},\hat{B'}]}}{2\left(\Delta\hat{B'}\right)^2}$. Notice that it corresponds indeed to a minimum, for the coefficient in front of $\lambda^2$ is positive in \eqref{mean value O} (which is basically a parabola).
We now apply our findings to $\hat{A}=\hat{X}$ and $\hat{B}=\hat{P}$. The minimum is then $\frac{\h}{2\left(\Delta\hat{P}\right)^2}$. In particular, one has that $\mean{\mathcal{O}(\lambda)} = 0$ if and only if \begin{align} \left(\hat{X'}+i\lambda_{min}\hat{P'}\right)\ket{\psi} &= 0\\ \left(X-\mean{X}\right)\ket{\psi} &= -\frac{i\h}{2\left(\Delta\hat{P}\right)^2}\left(\hat{P}-\mean{\hat{P}}\right)\ket{\psi} \end{align} Let us now write $l = \Delta\hat{X}$. Because we define a coherent state as the saturation of the Heisenberg uncertainty principle, we can write $2\left(\Delta\hat{P}\right)^2 = \frac{\h^2}{l^2}$. Hence, one has that \begin{equation} \left(\hat{X}-\mean{\hat{X}}\right)\ket{\psi} = -i\frac{l^2}{\h}\left(\hat{P}-\mean{\hat{P}}\right)\ket{\psi}\tag{2}\label{b wave function} \end{equation} Using the x-representation expression of $\hat{P}$ (see answers above for more details on this), we find the differential equation \begin{align} -x\psi(x)+\expval{\hat{X}}\psi(x)&=\frac{il^2}{\h}\left(\frac{\h}{i}\psi'(x)-\expval{\hat{P}}\psi(x)\right)\notag\\ \psi'(x) &= \left[\frac{\mean{\hat{X}}-x}{l^2}+\frac{i}{\h}\mean{\hat{P}}\right]\psi(x)\tag{3}\label{wave function} \end{align} The solution to this is given by $\psi(x) = \psi_0\exp(-\frac{\left(x-\mean{\hat{X}}\right)^2}{2l^2}+\frac{i}{\h}\mean{\hat{P}}x)$ where $\psi_0$ is a normalization constant.
Through an appropriate Gaussian integral, one finds that the wave function of a coherent state is given by \begin{equation} \psi(x) = \frac{\exp(-\frac{i}{2\pi}\mean{\hat{X}}\mean{\hat{P}})}{\left(\pi l^2\right)^{1/4}}\exp\left(-\frac{\left(x-\mean{\hat{X}}\right)^2}{2l^2}+\frac{i\mean{\hat{P}}x}{\hbar}\right)\tag{4}\label{wave function coherent states} \end{equation} \eqref{wave function coherent states} shows that the sought after wave function is a gaussian. Notice we added a global phase of $\exp(-\frac{i}{2\pi}\mean{\hat{X}}\mean{\hat{P}})$. Global phases are arbitrary and do not change the physics of the problem. An appropriate phase might be chosen in such a way as to simplify future considerations, and that is what is done here.
Interlude : other formulations of coherent states
I am going to develop here the mathematics behind coherent states, so as to find two other formulations of them that are easier to use. In particular, we're going to be able to use algebra instead of calculus - which is arguably (in my experience) simpler to manipulate. The reader only interested in the proposed alternative solution to the problem arised by the OP may skip to the next section.
Eigenvectors of the annihilation operator
Let us introduce $\hat{a} = \frac{1}{2}\left(\frac{\hat{X}}{l}+\frac{il\hat{P}}{\h}\right)$. One easily checks that $[\hat{a},\hat{a}^\dagger] = \mathbb{I}$. We call $\hat{a}$ the annihilation operator, whereas $\hat{a}^\dagger$ is known as the creation operator. The reasons behind these names are obvious when one studies the eigenvalues of $\hat{N} = \hat{a}^\dagger\hat{a}$.
Let $\ket{\psi}$ be an element of an Hilbert space $\mathcal{H}$. What are the solutions of the eigen-value problem $\hat{a}\ket{\psi} = \alpha\ket{\psi}$? Using the defintion of $\hat{a}$ provided, one has that $\frac{1}{2}\left(\frac{\hat{X}}{l}+\frac{il\hat{P}}{\h}\right)\ket{\psi} = \alpha\ket{\psi}$. One steadily finds that $\alpha = \frac{1}{2}\left(\frac{\mean{\hat{X}}}{l}+\frac{il\mean{\hat{P}}}{\h}\right)$. Putting this in the eigen-value problem, one finds (after a few appropriate manipulations) the equation \eqref{b wave function} again.
This means that the wave function defined by \eqref{wave function coherent states} corresponds to the eigenvectors of $\hat{a}$. Per convention, we write $\ket{\psi} = \ket{\alpha}$.
Theorem 3 $\ket{\alpha}$ is a coherent state (as defined in Definition 2) if and only if $\ket{\alpha}$ verifies \begin{equation} \hat{a}\ket{\alpha}=\alpha\ket{\alpha}\tag{5}\label{coherent in eigen} \end{equation} for any $\alpha\in\mathbb{C}$.
Displacement operator
We define the displacement operator as $\hat{D}(\alpha) = e^{\alpha\hat{a}^\dagger-\alpha^*\hat{a}}$.
Property 4 (Glauber) Let $\hat{A}$ and $\hat{B}$ be two operators. The relation $e^\hat{A}e^\hat{B} = e^{\hat{A}+\hat{B}}e^{-\frac{1}{2}[\hat{A},\hat{B}]}$ holds true if and only if ($\hat{A},\hat{B}$) commutes with [$\hat{A},\hat{B}$].
Through an appropriate use of Property 4, one shows the displacement operator defined above can equivalently be defined as $\hat{D}(\alpha) = e^{-\frac{1}{2}\mathbb{I}}e^{\alpha\hat{a}^\dagger}e^{-\alpha^*\hat{a}}$.
Property 5 One can show that a coherent state can be defined as the Displacement operator applied to the ground state $\ket{0}$: \begin{equation} \ket{\alpha} = \hat{D}(\alpha)\ket{0}\tag{6}\label{eq:coherent in displacement} \end{equation}
Here you go! Three equivalent formulations of coherent states. Of course, this was a very quick treatment and the reader interested in further developments in the topic may consult other sources. I suggest reading Prof. Reinhold's lecture notes on the matter.
Alternative solution to the question
Without any further mathematical developments or proofs, let us bring to the reader's attention (I may edit the post, later on, to add some maths to back up the following claim) that coherent states closely imitates the classical solutions of the harmonic oscillator. That means that if applied to the Hamiltonian of an harmonic oscillator, the means values of the position and impulsion operators follow the classical solution of the Harmonic Oscillator. The reader may read more about this here.
The OP did not specify, but I reckon he used the ground state in the Fock space $\ket{n}$. Although coherent states and Fock states are not the same, it turns out their ground states are equivalent. Moreover, one may prove the following relation \begin{equation} \ket{\alpha} = e^{-\frac{1}{2}\norm{\alpha}^2}\sum_n \frac{\alpha^n}{\sqrt{n!}}\ket{n} \end{equation} which would allow one to define a coherent state as a linear combination of Fock states. Using this, one may see that the ground state of a quantum harmonic oscillator is $\braket{x}{0} = \psi_0(x)$. Taking $\ket{0}$ to be the coherent void, one only has to put $\alpha = 0$ in \eqref{wave function coherent states}. This gives us an alternative way of computing of the ground state of the quantum harmonic oscillator!
PS: I don't have time to read everything I wrote now. There is probably lots of grammatical mistakes, and I hope as little as possible mathematical ones. Feel free to edit if it feels necessary.